Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-7x+4y &= -8 \\ 5x-6y &= 1\end{align*}$
Answer: Begin by moving the $y$ -term in the second equation to the right side of the equation. $5x = 6y+1$ Divide both sides by $5$ to isolate $x$ $x = {\dfrac{6}{5}y + \dfrac{1}{5}}$ Substitute this expression for $x$ in the first equation. $-7({\dfrac{6}{5}y + \dfrac{1}{5}}) + 4y = -8$ $-\dfrac{42}{5}y - \dfrac{7}{5} + 4y = -8$ Simplify by combining terms, then solve for $y$ $-\dfrac{22}{5}y - \dfrac{7}{5} = -8$ $-\dfrac{22}{5}y = -\dfrac{33}{5}$ $y = \dfrac{3}{2}$ Substitute $\dfrac{3}{2}$ for $y$ in the top equation. $-7x+4( \dfrac{3}{2}) = -8$ $-7x+6 = -8$ $-7x = -14$ $x = 2$ The solution is $\enspace x = 2, \enspace y = \dfrac{3}{2}$.